SRCC GBO 2024

Rama and Jaya differ by a percentage of 10 and marks of 8. Jaya's share is 45%, and 55% is Rama's. Each percent corresponds to 4/5 marks. From this, rams's mark is 44, and that of Jaya is 36.

The coefficient of the x term in a quadratic equation represents the sum of roots, while the constant term indicates the products of roots. Directly for the P(x) $$\alpha,\ \beta\ $$ are the roots. In q(x), the x coefficient can be written as $$\alpha\ +1+\beta\ +1$$. By this, we say $$\alpha\ +1,\beta\ \ +1$$ are the roots of q(x).

Now, check with the options where the root of p(x) is one less than q(x). It is option C.

The distance between the two opposite sides is 18 which is just the length of shorter diagonal. That is $$ \sqrt{\ 3}a$$. Hence a is $$6\sqrt{\ 3}$$.

The growth percentage in 11 years is $$(224500-150000)/150000$$. That is 49.6666. Divide it by 11 to get an average percentage. That is 4.52

This is an AP with a common difference of 0.5.Use the formulae of $$n^{th}\ term\ =\ a+\left(n-1\right)\cdot d$$.

$$-0.25+\left(n-1\right)\cdot0.5\ =\ 17.25$$

n will 36.

Slit the given numbers as follows.

321 - 107*3

48 - $$2^4$$*3

66 - 11*2*3

To have a perfect square, we need to have even powers to the prime numbers. That is $$2^4\cdot107^2\cdot11^2\cdot3^2$$. That will be $$16 \times 1089 \times 11449 $$.

There are four different equations possible from this given equation. They are 

x+y=2

x-y=2

-x+y=2

-x-y=2

These lines forms a rhombus having diogonal length 4. The area of rhombus is $$\frac{1}{2}\cdot d_1\cdot d_2$$.

$$\frac{1}{2}\cdot4\cdot4\ =\ 8$$

Lets expand the given equation.

$$91\left(1+\frac{1}{x}+\frac{1}{y}+\frac{1}{xy}\right)$$

$$91\left(1+\frac{x+y}{xy}+\frac{1}{xy}\right)$$

x+y is constant, 91 is constant.

To get the minimum value we have to minimixe the xy term.

That happens when both are equal.

So x=26, y=26.

$$91\left(\left(1+\frac{1}{26}\right)\left(1+\frac{1}{26}\right)\right)$$

On simiplifing we get $$\frac{5103}{52}$$

Here we need to determine to which quadrant each of the g1, g2, g3, g4 goes to. For that we need to find the range and domain of these functions. The domain of g1 is same as that of f as the value is directly inserted in to the function. Since the value of g1(x) is 2 less than the f(x) the range would be(-1,0). So g1 goes to 3rd quadrant. Similarly for g2 the domain should be negative of what f has as the input for f is